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C6L1V@xxxxxxx wrote:
> Ralf Goertz wrote:
> > C6L1V@xxxxxxx wrote:
> >
> > > Ralf Goertz wrote:
> > >
> > >> In general, alpha must be smaller than or equal to 2pi*(n-1)/n.
> > >
> > > This lookss wrong: alpha must be <= 2pi/n, because if alpha (the
> > > smallest angle) is > 2pi/n,
> > > the sum of all the angles is > 2pi. Anyway, as I said, Feller solves
> > > the problem.
> >
> > Thank you for your comments, I think I can now figure out the recursive
> > relation myself. There seems to be a misunderstanding though. alpha
> > denotes the smallest angle that will *cover* all points.
>
> My apologies: I mis-read the question. However, Feller also solves your
> actual problem, too, I think. It seem to me that the smallest angle
> that covers all the points must equal 2*pi minus the largest gap. The
> distribution function of the largest gap is given by Feller's formula
> (9.9) on page 28 (as remarked on page 29). For a circle of
> circumference b, the distribution P{largest gap <=y} is
> sum_k=0^n (-1)^k (n choose k) 1{ky/b <=1} (1 - ky/b)^(n-1)
> = 1 - n*1{y <= b} (1-y/b)^(n-1) + n(n-1)/2 * 1{y <= b/2} (1-2y/b)^(n-1)
> + ... .
> Does this seem OK to you?
>
Checks out for n = 2, 3 and 4 and looking good! It's pleasing that it
unravels into an expression with such relatively simple if/then
clauses. I must say I expected more of a tangle.
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